Question

The non - zero vectors $\stackrel{\rightharpoonup }{a},\stackrel{\rightharpoonup }{b}$ and $\stackrel{\rightharpoonup }{c}$ are related by $\stackrel{\rightharpoonup }{a}=8\stackrel{\rightharpoonup }{b}$ and $\stackrel{\rightharpoonup }{c}=-7\stackrel{\rightharpoonup }{b}$. Then, the angle between $\stackrel{\rightharpoonup }{a}$ and $\stackrel{\rightharpoonup }{c}$ is?

• A. $\mathrm{\pi }$
• B. $0$
• C. $\frac{\mathrm{\pi }}{4}$
• D. $\frac{\mathrm{\pi }}{2}$

Answer 1

The correct option is A

$\mathrm{\pi }$

Find the angle between the two given vectors

Consider the vector $\stackrel{\rightharpoonup }{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$

From given data vectors $\stackrel{\rightharpoonup }{a}$ and $\stackrel{\rightharpoonup }{c}$ can be written as

$\stackrel{\rightharpoonup }{a}=8\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\right)$

$\stackrel{\rightharpoonup }{c}=-7\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\right)$

Let the angle between $\stackrel{\rightharpoonup }{a}$ and $\stackrel{\rightharpoonup }{c}$ be $\theta$

We know that

$\cos \left(\theta \right)=\frac{\stackrel{\rightharpoonup }{a}.\stackrel{\rightharpoonup }{c}}{\left|\stackrel{\rightharpoonup }{a}\right|\left|\stackrel{\rightharpoonup }{c}\right|}$

$=\frac{8\times \left(-7\right)\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\right).\left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\right)}{8\sqrt{{b^2}_1+{b^2}_2+{b^2}_3}\times 7\sqrt{{b^2}_1+{b^2}_2+{b^2}_3}}$ $\left(\because \hat{i}.\hat{i}=\hat{j}.\hat{j}=\hat{k}.\hat{k}=1\right)$

$=\frac{-56\left({b^2}_1+{b^2}_2+{b^2}_3\right)}{56\left({b^2}_1+{b^2}_2+{b^2}_3\right)}$

$\Rightarrow$ $\cos \left(\theta \right)=-1$

$\Rightarrow$ $\theta ={\cos }^{-1}\left(-1\right)$

$\Rightarrow$ $\theta =\mathrm{\pi }$

Hence, correct option is $\left(A\right)$